Question

Suppose $a,b,c$ are in AP and $a^2,b^2,c^2$ are in GP, if $a<b<c$ and $a+b+c=$ $\frac{3}{2}$, then the value of $a$ is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{2}-\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Given:

a,b,c are in A.P.

$\therefore 2b=a+c$

$\therefore a+b+c=\frac{3}{2}$

$\Rightarrow 3b=\frac{3}{2}$
$\therefore b=\frac{1}{2}$

$\therefore a+c=1$ .....(1)

$\Rightarrow c=(1-a)$

$\therefore a^2{c}^2=b^4=\frac{1}{16}$ .....(2)

From eq (1) and (2):

$\Rightarrow a(1-a)=\pm \frac{1}{4}$

$\Rightarrow a^2-a\mp \frac{1}{4}=0$

$\Rightarrow a^2-a-\frac{1}{4}=0$ or $a^2-a+\frac{1}{4}=0$

$\Rightarrow a=\frac{1\pm \sqrt{1+1}}{2}$ or $(a-\frac{1}{2}{)}^2=0$

$\therefore a=(\frac{1}{2}+\frac{1}{\sqrt{2}}),(\frac{1}{2}-\frac{1}{\sqrt{2}}),\left(\frac{1}{2}\right)$

Given: $a<b<c$

$\therefore a=\frac{1}{2}-\frac{1}{\sqrt{2}}$

$(Ans\to D)$